Chapter 1
Graphs, Functions, and Models
To graph (−1, 4) we move from the origin 1 unit
Check Your Understanding Section 1.1 left of the y-axis. Then we move 4 units up from
x-axis.
1. The point ( —5, 0) is on an axis, so it is not in any quadrant. To graph (0, 2) we do not move to the right or the
The statement is false. the y-axis since the first coordinate is 0. From the
we move 2 units up.
2. The ordered pair (1,—6) is located 1 unit right of the origin
and 6 units below it. The ordered pair (—6, 1) is located 6 To graph (2, —2) we move from the origin 2 units
units left of the origin and 1 unit above it. Thus, (1,— 6) right of the y-axis. Then we move 2 units down fro
and (—6, 1) do not name the same point. The statement x-axis.
is false. y
3. True; the first coordinate of a point is also called the
( 1, 4) 4
abscissa.
4. True; the point ( 2 7) is 2 units left of the origin and 2 (0, 2)
−, (4, 0)
7 units above it. 4 2 2 4
2 (2, 2)
5. True; the second coordinate of a point is also called the ( 3, 5) 4
ordinate.
6. False; the point (0, −3) is on the y-axis. 5. To graph ( —5, 1) we move from the origin 5 units
left of the y-axis. Then we move 1 unit up from the x
To graph (5, 1) we move from the origin 5 units to th
Exercise Set 1.1 of the y-axis. Then we move 1 unit up from the x
To graph (2, 3) we move from the origin 2 units to th
1. Point A is located 5 units to the left of the y-axis and of the y-axis. Then we move 3 units up from the x
4 units up from the x-axis, so its coordinates are (−5, 4).
To graph (2, —1) we move from the origin 2 units
Point B is located 2 units to the right of the y-axis and right of the y-axis. Then we move 1 unit down fro
2 units down from the x-axis, so its coordinates are (2, −2). x-axis.
Point C is located 0 units to the right or left of the y-axis To graph (0, 1) we do not move to the right or the
and 5 units down from the x-axis, so its coordinates are the y-axis since the first coordinate is 0. From the
(0, −5). we move 1 unit up.
Point D is located 3 units to the right of the y-axis and
5 units up from the x-axis, so its coordinates are (3, 5). y
Point E is located 5 units to the left of the y-axis and 4
4 units down from the x-axis, so its coordinates are 2
(2, 3)
(−5, −4). ( 5, 1) (0, 1) (5, 1)
4 2
Point F is located 3 units to the right of the y-axis and 2 (2, 1)
0 units up or down from the x-axis, so its coordinates are
4
(3, 0).
3. To graph (4, 0) we move from the origin 4 units to the right
7. The first coordinate represents the year and the
of the y-axis. Since the second coordinate is 0, we do not
sponding second coordinate represents the number o
move up or down from the x-axis.
served by Southwest Airlines. The ordered pai
To graph (−3, −5) we move from the origin 3 units to the (1971, 3), (1981, 15), (1991, 32), (2001, 59), (2011
left of the y-axis. Then we move 5 units down from the and (2021, 121).
x-axis.
c 2025 Pearson Education, Inc.
Copyright ◯
,14 Chapter 1: Graphs, Functions, and Mo
9. To determine whether (−1, −9) is a solution, substitute 2a + 5b = 3
−1 for x and −9 for y. 3
y = 7x − 2 2·0+5· ? 3
5
−9 ?¯ 7(−1) − 2 0+3 ¯
¯ −7 − 2 3 ¯ 3 TRUE³
3´
−9 ¯ −9 TRUE The equation 3 = 3 is true, so 0, is a solution.
The equation −9 = −9 is true, so (−1, −9) is a solution. 5
To determine whether (0, 2) is a solution, substitute 0 for 15. To determine whether (−0.75, 2.75) is a solution, s
x and 2 for y. tute −0.75 for x and 2.75 for y.
y = 7x − 2 x2 − y2 = 3
2 ? 7·0−2 (−0.75)2 − (2.75)2 ?¯ 3
¯ 0— 2 0.5625 − 7.5625
¯
2 ¯ −2 FALSE −7 ¯ 3 FALSE
The equation 2 = −2 is false, so (0, 2) is not a solution. The equation −7 = 3 is false, so ( −0.75, 2.75) is
³ 2 3´ 2 solution.
11. To determine whether , is a solution, substitute To determine whether (2, −1) is a solution, substi
3 4 3
3 for x and −1 for y.
for x and for y.
4 x2 − y2 = 3
6x − 4y = 1
22 − (−1)2 ?¯ 3
2 3 4— 1
6· −4· ? 1 ¯
3 4 ¯
4 3 3 ¯ 3 TRUE
— The equation 3 = 3 is true, so (2, −1) is a solution.
1 ¯ 1 TRUE
³ 2 3´ 17. Graph 5x − 3y = −15.
The equation 1 = 1 is true, so , is a solution.
³ 3´ 3 4 xTo find the x-intercept we replace y with 0 and sol
To determine whether 1, is a solution, substitute 1 for .
2 5x − 3 · 0 = −15
3
x and for y. 5x = −15
2
x = −3
6x − 4y = 1
The x-intercept is (−3, 0).
3
6·1−4· ? 1 To find the y-intercept we replace x with 0 and sol
2
y.
6−6 ¯
5 · 0 − 3y = −15
0 ¯ 1 FALSE
³ 3´ −3y = −15
The equation 0 = 1 is false, so 1, is not a solution. y=5
2
³ 1 4´ The y-intercept is (0, 5).
13. To determine whether − , − is a solution, substitute
2 5 We plot the intercepts and draw the line that co
1 4 them. We could find a third point as a check th
− for a and − for b.
2 5 intercepts were found correctly.
2a + 5b = 3
³ 1´ ³ 4´ y
2 − +5 − ?3 (0, 5)
4
2 5
2 5x 3y 15
−1 − 4 ( 3, 0)
4 2 2 4 x
−5 ¯ 3 FALSE 2
³ 1 4´
The equation −5 = 3 is false, so − , − is not a solu- 4
2 5
tion.
³ 3´
To determine whether 0, is a solution, substitute 0 for
5
3
a and for b.
5
c 2025 Pearson Education, Inc.
Copyright ◯
,Exercise Set 1.1
19. Graph 2x + y = 4. When x = 0, y = 3x + 5 = 3 · 0 + 5 = 0 + 5 = 5
To find the x-intercept we replace y with 0 and solve for We list these points in a table, plot them, and dra
x. graph.
2x + 0 = 4 y
x y (x, y)
2x = 4 6
x=2 −3 −4 (−3, −4) y 3x 5
2
The x-intercept is (2, 0). −1 2 (−1, 2)
4 2 4
To find the y-intercept we replace x with 0 and solve for 0 5 (0, 5) 2
y.
2y2·y20y2+y2yy 2 = y24y2
25.y 2 Graphy2xy2−y2yy2=y23.
yy 2 = y24y2
Makey2ay2tabley2ofy2values,y2ploty2they2pointsy2iny
They 2 y-intercepty 2 isy 2 (0,y24).
table,y2andy2drawy2they2graph.
Wey2ploty2they2interceptsy2andy2drawy2they2liney2thaty2 y
containsy2them.y 2 Wey2couldy2findy2ay2thirdy2pointy2asy x y (x,y2y) y 2 xy 2 y
4
2ay2checky2thaty2they2interceptsy2werey2foundy2correctl 2 yy 2 y 2
y. −2 −5 (−2,y2−5) 2
0 −3 (0,y2−3) 4 2 2 4 x
y 2
2x y 4 4y 2 (0, 3 0 (3,y20) 4
y24)
2 (2,y2 3y2
4 2 0) 27.y 2 Graphy2 yy2 =y2−y2 xy2+y23.
4
2 2y 2
Byy2choosingy2multiplesy2ofy24y2fory2x,y2wey2cany2
4
y2fractiony2valuesy2fory2y.y 2 Makey2ay2tabley2ofy2v
2ploty2they2pointsy2iny2they2table,y2andy2drawy2th
21.y 2 Graphy24yy2−y23xy2=y212. ph.
y
Toy 2 findy 2 they 2 x- x y (x,y2y)
4
intercepty 2 wey 2 replacey 2 yy 2 withy 2 0y 2 andy 2 solvey 2 for −4 6 (−4,y2 6 2
x. )
4y2·y20y2−y23xy 2 =y 2 12 4 2 2y 2
0 3 (0,y23) 2 y2 4
−3xy 2 =y 2 12 y y2y 3
4 0 (4,y20)
4
2 34y2
xy2 =y2 −4
They 2 x-intercepty 2 isy 2 (−4,y20).
Toy2findy2they2y-
intercepty2wey2replacey2xy2withy20y2andy2solvey2fory2y. 29.y2 Graphy25xy2−y22yy2=y28.
4yy2−y23y2 ·y20 y2 =y 2 12 Wey2couldy2solvey2fory2yy
4yy 2 =y 2 12 2first.y25xy2−y22yy 2 = y 2 8
yy 2 = y23y2 y2
They 2 y-intercepty 2 isy 2 (0,y23). −2yy 2 =y2 −5xy2+y28y 2 y 2 Subtractingy2 5xy2 ony2 bot
es
Wey 2 ploty 2 they 2 interceptsy 2 andy 2 drawy 2 they 2 liney 5 1 ony 2 b
yy2 = x y2 −y 2 4
Multiplyingy 2 byy 2 −
2 thaty 2 contains
them.y 2 Wey 2 couldy 2 findy 2 ay 2 thirdy 2 pointy 2 asy 2 ay
2 checky 2 thaty 2 they2interceptsy2werey2foundy2correctly. 23.y 2 Graphy 2 yy2 =y2 3xy2+y25.
Wey 2 choosey 2 somey 2 valuesy 2 fory 2 xy 2 andy 2
y
hey 2 corresponding
y 2 4y 3x y-values.
12y 2
2y 2 (0,
y24
Wheny 2 xy2 =y2−3,y 2 yy2 =y23xy2+y25y2=y2 3
y23)
4 2 2y 2 y2 4 +y25y2=y2−9y2+y25y2=y2−4.
y
x
2
Wheny 2 xy2 =y2−1,y 2 yy2 =y23xy2+y25y2=y2 3
2( +y25y2=y2−3y2+y25y2=y22.
4
c 2025 Pearson Education, Inc.
Copyright ◯
, 16 Chapter 1: Graphs, Functions, and Mo
2 2
sides
Byy2choosingy2multiplesy2ofy22y2fory2xy2wey2cany2avoidy2fra
ctiony2valuesy2fory2y.y 2 Makey2ay2tabley2ofy2values,y2ploty2t
hey2pointsy2iny2they2table,y2andy2drawy2they2graph.
y
x y (x,y2y)
4
0 −4 (0,y2−4) 2
2 1 (2,y21) 4 2 2y2
2 y2 4
4 6 (4,y26) 4y 2
y2 5x 2y
8y 2
cy 2 2025y2Pearsony2Educati
Copyrighty2◯
Graphs, Functions, and Models
To graph (−1, 4) we move from the origin 1 unit
Check Your Understanding Section 1.1 left of the y-axis. Then we move 4 units up from
x-axis.
1. The point ( —5, 0) is on an axis, so it is not in any quadrant. To graph (0, 2) we do not move to the right or the
The statement is false. the y-axis since the first coordinate is 0. From the
we move 2 units up.
2. The ordered pair (1,—6) is located 1 unit right of the origin
and 6 units below it. The ordered pair (—6, 1) is located 6 To graph (2, —2) we move from the origin 2 units
units left of the origin and 1 unit above it. Thus, (1,— 6) right of the y-axis. Then we move 2 units down fro
and (—6, 1) do not name the same point. The statement x-axis.
is false. y
3. True; the first coordinate of a point is also called the
( 1, 4) 4
abscissa.
4. True; the point ( 2 7) is 2 units left of the origin and 2 (0, 2)
−, (4, 0)
7 units above it. 4 2 2 4
2 (2, 2)
5. True; the second coordinate of a point is also called the ( 3, 5) 4
ordinate.
6. False; the point (0, −3) is on the y-axis. 5. To graph ( —5, 1) we move from the origin 5 units
left of the y-axis. Then we move 1 unit up from the x
To graph (5, 1) we move from the origin 5 units to th
Exercise Set 1.1 of the y-axis. Then we move 1 unit up from the x
To graph (2, 3) we move from the origin 2 units to th
1. Point A is located 5 units to the left of the y-axis and of the y-axis. Then we move 3 units up from the x
4 units up from the x-axis, so its coordinates are (−5, 4).
To graph (2, —1) we move from the origin 2 units
Point B is located 2 units to the right of the y-axis and right of the y-axis. Then we move 1 unit down fro
2 units down from the x-axis, so its coordinates are (2, −2). x-axis.
Point C is located 0 units to the right or left of the y-axis To graph (0, 1) we do not move to the right or the
and 5 units down from the x-axis, so its coordinates are the y-axis since the first coordinate is 0. From the
(0, −5). we move 1 unit up.
Point D is located 3 units to the right of the y-axis and
5 units up from the x-axis, so its coordinates are (3, 5). y
Point E is located 5 units to the left of the y-axis and 4
4 units down from the x-axis, so its coordinates are 2
(2, 3)
(−5, −4). ( 5, 1) (0, 1) (5, 1)
4 2
Point F is located 3 units to the right of the y-axis and 2 (2, 1)
0 units up or down from the x-axis, so its coordinates are
4
(3, 0).
3. To graph (4, 0) we move from the origin 4 units to the right
7. The first coordinate represents the year and the
of the y-axis. Since the second coordinate is 0, we do not
sponding second coordinate represents the number o
move up or down from the x-axis.
served by Southwest Airlines. The ordered pai
To graph (−3, −5) we move from the origin 3 units to the (1971, 3), (1981, 15), (1991, 32), (2001, 59), (2011
left of the y-axis. Then we move 5 units down from the and (2021, 121).
x-axis.
c 2025 Pearson Education, Inc.
Copyright ◯
,14 Chapter 1: Graphs, Functions, and Mo
9. To determine whether (−1, −9) is a solution, substitute 2a + 5b = 3
−1 for x and −9 for y. 3
y = 7x − 2 2·0+5· ? 3
5
−9 ?¯ 7(−1) − 2 0+3 ¯
¯ −7 − 2 3 ¯ 3 TRUE³
3´
−9 ¯ −9 TRUE The equation 3 = 3 is true, so 0, is a solution.
The equation −9 = −9 is true, so (−1, −9) is a solution. 5
To determine whether (0, 2) is a solution, substitute 0 for 15. To determine whether (−0.75, 2.75) is a solution, s
x and 2 for y. tute −0.75 for x and 2.75 for y.
y = 7x − 2 x2 − y2 = 3
2 ? 7·0−2 (−0.75)2 − (2.75)2 ?¯ 3
¯ 0— 2 0.5625 − 7.5625
¯
2 ¯ −2 FALSE −7 ¯ 3 FALSE
The equation 2 = −2 is false, so (0, 2) is not a solution. The equation −7 = 3 is false, so ( −0.75, 2.75) is
³ 2 3´ 2 solution.
11. To determine whether , is a solution, substitute To determine whether (2, −1) is a solution, substi
3 4 3
3 for x and −1 for y.
for x and for y.
4 x2 − y2 = 3
6x − 4y = 1
22 − (−1)2 ?¯ 3
2 3 4— 1
6· −4· ? 1 ¯
3 4 ¯
4 3 3 ¯ 3 TRUE
— The equation 3 = 3 is true, so (2, −1) is a solution.
1 ¯ 1 TRUE
³ 2 3´ 17. Graph 5x − 3y = −15.
The equation 1 = 1 is true, so , is a solution.
³ 3´ 3 4 xTo find the x-intercept we replace y with 0 and sol
To determine whether 1, is a solution, substitute 1 for .
2 5x − 3 · 0 = −15
3
x and for y. 5x = −15
2
x = −3
6x − 4y = 1
The x-intercept is (−3, 0).
3
6·1−4· ? 1 To find the y-intercept we replace x with 0 and sol
2
y.
6−6 ¯
5 · 0 − 3y = −15
0 ¯ 1 FALSE
³ 3´ −3y = −15
The equation 0 = 1 is false, so 1, is not a solution. y=5
2
³ 1 4´ The y-intercept is (0, 5).
13. To determine whether − , − is a solution, substitute
2 5 We plot the intercepts and draw the line that co
1 4 them. We could find a third point as a check th
− for a and − for b.
2 5 intercepts were found correctly.
2a + 5b = 3
³ 1´ ³ 4´ y
2 − +5 − ?3 (0, 5)
4
2 5
2 5x 3y 15
−1 − 4 ( 3, 0)
4 2 2 4 x
−5 ¯ 3 FALSE 2
³ 1 4´
The equation −5 = 3 is false, so − , − is not a solu- 4
2 5
tion.
³ 3´
To determine whether 0, is a solution, substitute 0 for
5
3
a and for b.
5
c 2025 Pearson Education, Inc.
Copyright ◯
,Exercise Set 1.1
19. Graph 2x + y = 4. When x = 0, y = 3x + 5 = 3 · 0 + 5 = 0 + 5 = 5
To find the x-intercept we replace y with 0 and solve for We list these points in a table, plot them, and dra
x. graph.
2x + 0 = 4 y
x y (x, y)
2x = 4 6
x=2 −3 −4 (−3, −4) y 3x 5
2
The x-intercept is (2, 0). −1 2 (−1, 2)
4 2 4
To find the y-intercept we replace x with 0 and solve for 0 5 (0, 5) 2
y.
2y2·y20y2+y2yy 2 = y24y2
25.y 2 Graphy2xy2−y2yy2=y23.
yy 2 = y24y2
Makey2ay2tabley2ofy2values,y2ploty2they2pointsy2iny
They 2 y-intercepty 2 isy 2 (0,y24).
table,y2andy2drawy2they2graph.
Wey2ploty2they2interceptsy2andy2drawy2they2liney2thaty2 y
containsy2them.y 2 Wey2couldy2findy2ay2thirdy2pointy2asy x y (x,y2y) y 2 xy 2 y
4
2ay2checky2thaty2they2interceptsy2werey2foundy2correctl 2 yy 2 y 2
y. −2 −5 (−2,y2−5) 2
0 −3 (0,y2−3) 4 2 2 4 x
y 2
2x y 4 4y 2 (0, 3 0 (3,y20) 4
y24)
2 (2,y2 3y2
4 2 0) 27.y 2 Graphy2 yy2 =y2−y2 xy2+y23.
4
2 2y 2
Byy2choosingy2multiplesy2ofy24y2fory2x,y2wey2cany2
4
y2fractiony2valuesy2fory2y.y 2 Makey2ay2tabley2ofy2v
2ploty2they2pointsy2iny2they2table,y2andy2drawy2th
21.y 2 Graphy24yy2−y23xy2=y212. ph.
y
Toy 2 findy 2 they 2 x- x y (x,y2y)
4
intercepty 2 wey 2 replacey 2 yy 2 withy 2 0y 2 andy 2 solvey 2 for −4 6 (−4,y2 6 2
x. )
4y2·y20y2−y23xy 2 =y 2 12 4 2 2y 2
0 3 (0,y23) 2 y2 4
−3xy 2 =y 2 12 y y2y 3
4 0 (4,y20)
4
2 34y2
xy2 =y2 −4
They 2 x-intercepty 2 isy 2 (−4,y20).
Toy2findy2they2y-
intercepty2wey2replacey2xy2withy20y2andy2solvey2fory2y. 29.y2 Graphy25xy2−y22yy2=y28.
4yy2−y23y2 ·y20 y2 =y 2 12 Wey2couldy2solvey2fory2yy
4yy 2 =y 2 12 2first.y25xy2−y22yy 2 = y 2 8
yy 2 = y23y2 y2
They 2 y-intercepty 2 isy 2 (0,y23). −2yy 2 =y2 −5xy2+y28y 2 y 2 Subtractingy2 5xy2 ony2 bot
es
Wey 2 ploty 2 they 2 interceptsy 2 andy 2 drawy 2 they 2 liney 5 1 ony 2 b
yy2 = x y2 −y 2 4
Multiplyingy 2 byy 2 −
2 thaty 2 contains
them.y 2 Wey 2 couldy 2 findy 2 ay 2 thirdy 2 pointy 2 asy 2 ay
2 checky 2 thaty 2 they2interceptsy2werey2foundy2correctly. 23.y 2 Graphy 2 yy2 =y2 3xy2+y25.
Wey 2 choosey 2 somey 2 valuesy 2 fory 2 xy 2 andy 2
y
hey 2 corresponding
y 2 4y 3x y-values.
12y 2
2y 2 (0,
y24
Wheny 2 xy2 =y2−3,y 2 yy2 =y23xy2+y25y2=y2 3
y23)
4 2 2y 2 y2 4 +y25y2=y2−9y2+y25y2=y2−4.
y
x
2
Wheny 2 xy2 =y2−1,y 2 yy2 =y23xy2+y25y2=y2 3
2( +y25y2=y2−3y2+y25y2=y22.
4
c 2025 Pearson Education, Inc.
Copyright ◯
, 16 Chapter 1: Graphs, Functions, and Mo
2 2
sides
Byy2choosingy2multiplesy2ofy22y2fory2xy2wey2cany2avoidy2fra
ctiony2valuesy2fory2y.y 2 Makey2ay2tabley2ofy2values,y2ploty2t
hey2pointsy2iny2they2table,y2andy2drawy2they2graph.
y
x y (x,y2y)
4
0 −4 (0,y2−4) 2
2 1 (2,y21) 4 2 2y2
2 y2 4
4 6 (4,y26) 4y 2
y2 5x 2y
8y 2
cy 2 2025y2Pearsony2Educati
Copyrighty2◯
